Proof: Rearrange Sum 6 2
Let's prove the following theorem:
((((a + b) + c) + d) + e) + f = ((a + b) + (c + d)) + (e + f)
Proof:
# | Claim | Reason |
---|---|---|
1 | ((((a + b) + c) + d) + e) + f = (((a + b) + c) + d) + (e + f) | ((((a + b) + c) + d) + e) + f = (((a + b) + c) + d) + (e + f) |
2 | ((a + b) + c) + d = (a + b) + (c + d) | ((a + b) + c) + d = (a + b) + (c + d) |
3 | (((a + b) + c) + d) + (e + f) = ((a + b) + (c + d)) + (e + f) | if ((a + b) + c) + d = (a + b) + (c + d), then (((a + b) + c) + d) + (e + f) = ((a + b) + (c + d)) + (e + f) |
4 | ((((a + b) + c) + d) + e) + f = ((a + b) + (c + d)) + (e + f) | if ((((a + b) + c) + d) + e) + f = (((a + b) + c) + d) + (e + f) and (((a + b) + c) + d) + (e + f) = ((a + b) + (c + d)) + (e + f), then ((((a + b) + c) + d) + e) + f = ((a + b) + (c + d)) + (e + f) |
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